Question

If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches 0.200 atm and the temperature reaches 210 K, calculate the work done on the gas, the internalenergy change, and the heat absorbed by the gas.

Short answer

The work done on the gas is -323 J, the internal energy change is -393 J, and the heat absorbed by the gas is -70 J.

Step-by-step solution

First law of thermodynamics.

The first law of thermodynamics explains the different energy forms and energy interactions occurring in a particular system. The mathematical formula of this law is:

$\mathbf{\Delta U}\mathbf{=}\mathbf{q}\mathbf{+}\mathbf{w}$

The work done on the gas.

The value of is 0.1 atm.

The initial volume of the neon gas,${\text{V}}_{\text{1}}$can be given as:

${\text{V}}_{\text{1}}\text{=}\frac{{\text{nRT}}_{\text{1}}}{{\text{P}}_{\text{1}}}$

Here, n denotes the number of moles of gas, R is the gas constant, T is the temperature.

Substituting the values in the equation,

$$\begin{gathered} {\text{V}}_{\text{1}}\text{=}\frac{{\text{nRT}}_{\text{1}}}{{\text{P}}_{\text{1}}} \\ \text{=}\frac{{\text{0.5 mol×0.08206 L atmK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}\text{×273 K}}{\text{1.00 atm}} \\ \text{= 11.2 L} \end{gathered}$$

The final volume,${\text{V}}_{\text{2}}$ can be found as:

$$\begin{gathered} {\text{V}}_2\text{=}\frac{{\text{nRT}}_2}{{\text{P}}_2} \\ \text{=}\frac{{\text{0.5 mol×0.08206 L atmK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}\text{×210 K}}{\text{0.200 atm}} \\ \text{= 43.1 L} \end{gathered}$$

Thework done can be found as:

$$\begin{gathered} {\text{W = -P}}_{\text{ext}}\text{∆V} \\ \text{= - 0.1 atm×}\left({\text{V}}_{\text{2}}{\text{-V}}_{\text{1}}\right) \\ \text{= - 0.1 atm×}\left(\text{43.1 - 11.2}\right)\text{L×}\frac{\text{101.325 J}}{\text{1L atm}} \\ \text{= - 323 J} \end{gathered}$$

The internal energy change

Theinternal energy can be found as:

${\text{∆U = nC}}_{\text{v}}\text{∆T}$

Here, n is the number of moles,${\text{C}}_{\text{v}}$is specific heat at constant volume and$∆T$is the change in temperature.

The value of${\text{C}}_{\text{v}}$is:

$$\begin{gathered} {\text{C}}_{\text{v}}\text{=}\frac{\text{3}}{\text{2}}\text{R} \\ {\text{= 12.47JK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}} \end{gathered}$$

Therefore, the value of $∆U$ is:

$$\begin{gathered} \text{∆U =}\left(\text{0.5mol}\right)\left({\text{12.47JK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}\right)\left(\text{210 K - 273 K}\right) \\ \text{= - 393 K} \end{gathered}$$width="290">∆U =0.5mol×12.47JK- 1mol- 1×210 K - 273 K= - 393 K

The heat absorbed by the gas.

$\text{∆U = q + w}$

The value of q can be obtained as:

$$\begin{gathered} \text{q =}\text{U∆- w} \\ \text{= - 393 J -}\left(\text{- 323 J}\right) \\ \text{= - 70 J} \end{gathered}$$